Definition of regular vector of a Lie algebra

Question

I've seen two definitions for the regular vector of a Lie algebra $\mathfrak{g}$ of a compact Lie group and I wonder if they are equivalent. Let $\mathfrak{h}$ be a Cartan subalgebra of $\mathfrak{g}$, that is, the Lie algebra of a maximal torus in $G$.

1. We say that $X\in \mathfrak{h}$ is regular if $\mathfrak{z}(X)=\mathfrak{h}$; that is, $X$ commutes with only the elements of $\mathfrak{h}$.

The second definition is based on the notion of roots. Suppose $\mathfrak{g}=\mathfrak{h}\oplus \mathfrak{l}_1\oplus\dots \oplus \mathfrak{l}_k$ where $\mathfrak{l}_i$ are two-dimensional $ad_{\mathfrak{h}}$ invariant subspaces of $\mathfrak{g}$ with roots $\alpha_1, \dots, \alpha_k$.

2. We say that $X\in \mathfrak{h}$ is regular if $\alpha_1^2(X), \dots, \alpha_k^2(X)$ are nonzero distinct numbers.

I can see why $(2)\implies (1)$. This is because $ad_X$ is diagonalizable and hence $\dim \ker ad_X$ equals the multiplicity of the zero eigenvalue namely $\dim \mathfrak{h}$. But clearly $\mathfrak{z}(X)\supset\mathfrak{h}$, therefore $\mathfrak{z}(X)=\mathfrak{h}$.

Can anyone give me a hint on how to get the other implication? I know that $\alpha_i(X)^2$ are the eigenvalues of $ad_X^2$, but I can't quite see why this should imply the requirement about $\alpha_i(X)^2$. Are the two definitions equivalent?

Answer 1

If I understand correctly, the centralizer of $X\in\mathfrak{h}$ is $\mathfrak{h}\oplus\bigoplus_{i:\alpha_i(X)=0}\mathfrak{l}_i$. This centralizer is reduced to $\mathfrak{h}$ iff $X\notin\bigcup_i\mathrm{Ker}(\alpha_i)$. So this yields the equivalence if "distinct" is removed from 2.

Next one has to understand the condition that the $\alpha_i(X)^2$ are nonzero and distinct; this requires some more specific knowledge about roots.

First, your definition of root only defines $\alpha_i$ up to sign, but it does not matter since you square it. Then, to use the general theory, you need to relate this to roots in the usual sense. Then the roots $\beta_i$ in the complexification are precisely the $\pm i\alpha_i$.

So the condition is that $\beta_i(X)\neq 0$, $\beta_i(X)\pm\beta_j(X)\neq 0$ for all $i\neq j$. But looking at simple root systems (e.g., $A_1\times A_1$ or $A_2$) shows that we can find roots $\beta_i,\beta_j$ such that $\beta_i+\beta_j$ is not collinear to any root. So we can find $X$ such that $\alpha_i(X)^2=\alpha_j(X)^2$ although (1) holds. In conclusion (2) is in general stronger than (1).